Prove that a function that satisfies $f(x^{2^{n}})=f(x)^{2^{n}}$ is constant given $f(0)=1$ This question follows from this link: Entire functions such that $f(z^{2})=f(z)^{2}$
If $f(0)=1$, how can you conclude that because $f(z^{2^{n}})=f(z)^{2^{n}}$ for all natural $n$, the function $f$ is constant? Thanks in advance.
 A: Assuming that $f$ is entire, we can argue as follows:
Let $z$ be any complex number such that $|z| < 1$. Then we have that
$$
  \lim_{n \to \infty} z^{2^n} = 0,
$$
and since $f$ is continuous, we have that
$$
  \lim_{n \to \infty} f\left( z^{2^n} \right) = f(0) = 1.
$$
This implies that
$$
  \lim_{n \to \infty} f(z)^{2^n} = 1.
$$
Now if $|f(z)| < 1$, then we would have that
$$
  \lim_{n \to \infty} f(z)^{2^n} = 0
$$
and if $|f(z)| > 1$, we would have that
$$
  \lim_{n \to \infty} \left|f(z)^{2^n}\right| = \infty.
$$
We see that we must have that $|f(z)| = 1$ for every $z$ in the open unit ball. But it is a well known fact that if $|f|$ is constant on some domain, then so $f$. (It follows from the Cauchy-Riemann equations, for example.) Thus we have that $f(z) = 1$ for all $z$ such that $|z| < 1$. From this it readily follows that $f$ is constant on all of $\mathbb{C}$, since the zeroes of the function $z \mapsto f(z) - 1$ have a limit point, and hence the function must be identically $0$, which implies that $f$ is identically equal to $1$.
